Help with Titration lab needed

[engineer2010]

Homework Statement

A sol;n of 14.05 mL of NaOH is mixed with 50.0 mL of 0.2102 M H2SO4 (standardized) and allowed to react. The sol'n (H2SO4 and NaOH) is still acidic, so it is titrated with 10.55mL of standardized KOH (0.08883M) until the equilivance point is reached. What was the original concentration (molarity) of NaOH.

Homework Equations

I have converted H2SO4 to moles.
.2102 mol/L * .0500L = 0.01051 moles H2SO4

0.08883 mol/L KOH * 0.01055L = 0.0009372 mol KOH

The Attempt at a Solution

I am not sure how to link the 3 together

 

[symbolipoint]

Show us the reaction you are using for the neutralizations.

Basically (not jokingly), you are using two different bases to neutralize the sulfuric acid. Some moles of base are from KOH, and some unknown but calculable moles are from NaOH. HOW MANY BASE MOLES ARE NEEDED FOR THE NEUTRALIZATION? THis becomes just accounting.

 

[Blueshift5]

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I am happy to assist you with your titration lab. Based on the information provided, we can use the equation M1V1 = M2V2 to find the original concentration of NaOH. M1 and V1 represent the initial concentration and volume of NaOH, while M2 and V2 represent the final concentration and volume of NaOH at the equivalence point.

Using this equation, we can set up the following calculation:

M1 * 14.05 mL = 0.08883 mol/L * 10.55 mL

Solving for M1, we get:

M1 = (0.08883 mol/L * 10.55 mL)/14.05 mL = 0.0665 mol/L

Therefore, the original concentration of NaOH was 0.0665 mol/L. I hope this helps with your titration lab. Remember to always double check your calculations and units to ensure accuracy. Good luck!